A seocnd pendulum that gives correct time on earth, when shifted to mooon `(g_(m)=(1)/(6)g_(e))` the time loss in 1 min by second pendulum will be

Options

A. 10 s
B. 87 s
C. 60s
D. 30 s

Answer

B

Explanation

For earth `2=2pisqrt((I)/(g_(e)))`
For moon `T=2pi sqrt((I)/(g_(m)))`
`(2)/(T)=sqrt((g_(m))/(g_(e)))=(1)/(sqrt(6))`
`:. T= 2 sqrt(6)=4.9s`
For each oscilation loss `=4.9-2=2.9s`
In one minute pendulum should make 30 oscillations.
Hence, loss of time `=30xx2.9=87s`